Answer:
100
Step-by-step explanation:
We have the sum of first n terms of an AP,
Sn = n/2 [2a+(n−1)d]
Given,
36= 6/2 [2a+(6−1)d]
12=2a+5d ---------(1)
256= 16/2 [2a+(16−1)d]
32=2a+15d ---------(2)
Subtracting, (1) from (2)
32−12=2a+15d−(2a+5d)
20=10d ⟹d=2
Substituting for d in (1),
12=2a+5(2)=2(a+5)
6=a+5 ⟹a=1
∴ The sum of first 10 terms of an AP,
S10 = 10/2 [2(1)+(10−1)2]
S10 =5[2+18]
S10 =100
This is the sum of the first 10 terms.
Hope it will help.
Answer:
is one to one mapping, it is not onto mapping
Step-by-step explanation:
f₁(x) is one to one mapping
Let 
f₁(x) = f₁(y):
x₁³ = y₁³
f₁(x) is not onto mapping
Example: If f₁(x) = 7,
x₁³ = 7
![x_{1} = \sqrt[3]{7}](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20%5Csqrt%5B3%5D%7B7%7D)
x₁ is not an element of Z
is one to one mapping, it is not onto mapping
Answer:
2
Step-by-step explanation:
14 divided by 6 = 2.33333333
only 6 in a group so you round it
and you get 2
